Chapter 7.7, Problem 14P

To determine

To find the dimensions of the parking lot of greatest area.

Answer to Problem 14P

The dimensions of the parking lot of greatest area are 78 m by 78 m.

Explanation of Solution

Given information:

A parking lot is formed by fencing in a rectangular plot of land except for an entrance 12 m wide along one of the sides and 300 m of fencing is used to form the parking lot.

Let, the width and length of the rectangular plot are $x$ m and $y$ m respectively. The entrance is on the side of $x$ .

According to the question,

  $\begin{array}{l}x+(x-12)+2y=300\\ 2x-12+2y=300\\ 2x+2y=312\\ 2\cdot \left(x+y\right)=312\\ x+y=\frac{312}{2}=156\\ y=156-x\end{array}$

So, the length of the rectangular parking plot is $\left(156-x\right)$ m.

  $\therefore$ Area of the rectangular parking plot, $\text{A}$$=x\cdot y=x\cdot \left(156-x\right)=\left(156x-x^2\right)$ m²

The standard form of a quadratic equation: $y=a{x}^2+bx+c$ .

So, the formula of the axis of symmetry: $x=\frac{-b}{2a}$ .

Now, comparing the area of the parking plot with the standard form of the quadratic equation:

  $a=-1,b=156\text{ and }c=0$ .

By applying the formula of the axis of symmetry:

  $x=\frac{-b}{2a}=\frac{-156}{2\cdot (-1)}=\frac{156}{2}=78$ .

  $\therefore$ The width of the rectangular parking plot is 78 m.

Now, the length of the parking plot, $y=156-x=156-78=78$ m.

  $\therefore$ Greatest area of the parking plot $=156x-x^2=156\times 78-{\left(78\right)}^2=12168-6084=6084$ m².

Hence,

The dimensions of the parking lot of greatest area are 78 m by 78 m.